Explanation

(c)
Let $I = \int_{0}^{1}{\frac{dx}{x + \sqrt{x}} =}\int_{0}^{1}\frac{dx}{\sqrt{x}(\sqrt{x} + 1)}$
Put $\sqrt{x} = t \Rightarrow \frac{1}{2\sqrt{x}}dx = dt$
$\Rightarrow \frac{1}{\sqrt{x}}dx = 2dt$
$I = \int_{0}^{1}{\frac{2dt}{t + 1} = 2\left\lbrack \log(t + 1) \right\rbrack_{0}^{1}}$
$= 2(\log{2 - \log{0) = \log 4}}$